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son4ous [18]
2 years ago
9

Please help meee<3~I have no idea what the answer is~

Physics
1 answer:
valina [46]2 years ago
4 0
9).
In a properly conducted experiment, the experimenter controls one part
of the experiment to see what the other parts do.

Example:  Experiment to describe the effect of heat on ice.
Take two same-size ice cubes out of the same ice tray in the same fridge.
Place each one on a little temperature-controlled electric pad.
Turn one pad on, to make it warm.  Leave the other pad turned off. 
You CONTROL one part of the experiment:  the amount of heat that
       the ice cube gets.
You KNOW that the heat is the only thing different between the two
     ice cubes.  They're the same size.  They were both made from
     the same water, and froze in the same tray in the same fridge.
so
You KNOW that any difference will be the result of the heat on one of them.
You WATCH to see what happens to the one that gets the heat.


10).
An hypothesis is a prediction of what you believe may be true. 
Once you have it, it's time to do an experiment to find out whether
your hypothesis is true.

Example:
I have an hypothesis.  It predicts that when ice gets warm it melts.
Experiment:
Take two same-size ice cubes out of the same ice tray in the same fridge.
Set one ice cube down on the table.
Keep the other one in your hand.
The one in your hand melts while the one on the table is still solid.
Is the hypothesis correct ?
Maybe it is.  Maybe it isn't.
We know that there's something about your hand that melts ice.
It may be the warmth.  But it may be something else about human skin.
You'll need another experiment, slightly different, to find out if it's the warmth.

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An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which
Irina-Kira [14]

Answer:

2\times 10^{-3}\ C

6000

1.2 J

3.33\times 10^{-6}\ F

Explanation:

I = Current = 1 A

t = Time = 2 ms

n = Number of electrocyte

V = Voltage = 100 mV

Charge is given by

Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C

The charge flowing through the electrocytes in that amount of time is 2\times 10^{-3}\ C

The maximum potential is given by

V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000

The number of electrolytes is 6000

Energy is given by

E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F

The equivalent capacitance of all the electrocyte cells in the electric eel is 3.33\times 10^{-6}\ F

8 0
2 years ago
Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights
Arlecino [84]

Answer:

               275 kPa

Explanation:

             mass of the gas=m=1.5 kg

             initial volume if the gas=V₁=0.04 m³

             initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

             V₂=final volume

             V₂=2 V₁

As the temperature is constant

             T₁=T₂=T

\frac{P1V1}{T1}=\frac{P2 V2}{T2}

putting the values in the equation.

\frac{P1V1}{T1}=\frac{P2 *2V1}{T2}

P₂=\frac{P1}{2}

P₂=\frac{550}{2}

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

           

3 0
2 years ago
For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

no idea

Explanation:

7 0
2 years ago
A man swims at a speed of 0.4 m/s. How long will it take him to cross a pool of length 50 m?
konstantin123 [22]
Distance = speed * time, then time = distance / speed.

time = 50 / 0.4 = 125 s
7 0
2 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
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